Question

Let $y=y\left(x\right)$ be the solution of the differential equation $(y+1){\tan }^{2}xdx+\tan xdy+ydx=0,x\in (0,\frac{\pi }{2}).$ If $\underset{x\to 0^{-}}{lim}xy\left(x\right)=1,$ then the value of $y\left(\frac{\pi }{4}\right)$ is

• A. $\frac{\pi }{4}+1$
• B. $-\frac{\pi }{4}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{4}-1$

Answer 1

The correct option is C $\frac{\pi }{4}$

$(y+1){\tan }^{2}xdx+\tan xdy+ydx=0$
$\Rightarrow \frac{dy}{dx}=\frac{(y+1){\tan }^{2}x+y}{-\tan x}$
$\Rightarrow \frac{dy}{dx}=-(y+1)\tan x-y\cot x$
$\Rightarrow \frac{dy}{dx}+y(\tan x+\cot x)=-\tan x$
$I.F.=e^{\int (\tan x+\cot x)dx}=e^{\int \frac{{\sec }^{2}x}{\tan x}dx}$
$\therefore I.F.=\tan x$
$\Rightarrow y\tan x=\int -{\tan }^{2}xdx$
$\Rightarrow y\tan x=\int (1-{\sec }^{2}x)dx$
$\Rightarrow y\tan x=x-\tan x+c$

$\underset{x\to 0^{-}}{lim}xy\left(x\right)=1$
$\Rightarrow \underset{x\to 0^{-}}{lim}\frac{x}{\tan x}(x-\tan x+c)=1$
$\Rightarrow c=1$
$\therefore y\left(\frac{\pi }{4}\right)=\frac{\pi }{4}$